| ⇦ | 
| ⇨ |
Two springs with spring constants K₁=1500 N/m and K₂=3000 N/m are stretched by the same force. The ratio of potential energy stored in springs will be
Options
(a) 1:2
(b) 2:1
(c) 4:1
(d) 1:4
Correct Answer:
2:1
Explanation:
Force F=-Kxdisplacement (x),
Potential Energy = U =(1/2) Kx²
U = (1/2) K(F/K)² = (1/2)(F²/K)
U₁/U₂=(1/2)(F²/K₁)x(2K₂/F²)
K₂/K₁=3000/1500=2/1.
Related Questions: - For an ideal gas of diatomic molecules
- If the value of potential in an A.C circuit is 10 V, the peak value of potential is
- The electron drift speed is small and the change of the electron is also small
- In a radioactive decay process, the negatively charged emitted β- particles are
- Two resistors of 6Ω and 9Ω are connected in series to a 120 V source.
Topics: Laws of Motion
(103)
Subject: Physics
(2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- For an ideal gas of diatomic molecules
- If the value of potential in an A.C circuit is 10 V, the peak value of potential is
- The electron drift speed is small and the change of the electron is also small
- In a radioactive decay process, the negatively charged emitted β- particles are
- Two resistors of 6Ω and 9Ω are connected in series to a 120 V source.
Topics: Laws of Motion (103)
Subject: Physics (2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
18000+ students are using NEETLab to improve their score. What about you?
Solve Previous Year MCQs, Mock Tests, Topicwise Practice Tests, Identify Weak Topics, Formula Flash cards and much more is available in NEETLab Android App to improve your NEET score.
Share this page with your friends
Leave a Reply